Feb 4, 2018 · The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f …

Dec 15, 2017 · A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path integrals do not seek to …

Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

@user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the …

Nov 29, 2013 · Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". This MIT page says, "The more common name for the antiderivative is the indefinite integral." One is free …

Feb 17, 2025 · The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression and use …

The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions …

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

One possible interpretation: a "normal" integral is simply a line integral where the path is straight and oriented along a particular axis. Thus, as soon as you perform a transformation to the integrand to …

What is the integral of $$\int_ {-\infty}^ {\infty}e^ {-x^2/2}dx\,?$$ My working is here: = $-e^ (-1/2x^2)/x$ from negative infinity to infinity. What is the value of this?